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Complete the table of values for f(x)=2^x–2 and g(x)=x+4.

x f(x) g(x)
3
4
5
6
Both f(x) and g(x) grow as x gets larger and larger. Which function eventually exceeds the other?

1 answer

Let's calculate the values of \( f(x) = 2^x - 2 \) and \( g(x) = x + 4 \) for \( x = 3, 4, 5, \) and \( 6 \).

Calculating \( f(x) \) and \( g(x) \):

  1. For \( x = 3 \):

    • \( f(3) = 2^3 - 2 = 8 - 2 = 6 \)
    • \( g(3) = 3 + 4 = 7 \)

  2. For \( x = 4 \):

    • \( f(4) = 2^4 - 2 = 16 - 2 = 14 \)
    • \( g(4) = 4 + 4 = 8 \)

  3. For \( x = 5 \):

    • \( f(5) = 2^5 - 2 = 32 - 2 = 30 \)
    • \( g(5) = 5 + 4 = 9 \)

  4. For \( x = 6 \):

    • \( f(6) = 2^6 - 2 = 64 - 2 = 62 \)
    • \( g(6) = 6 + 4 = 10 \)

Complete Table of Values:

\[ \begin{array}{|c|c|c|} \hline x & f(x) & g(x) \ \hline 3 & 6 & 7 \ 4 & 14 & 8 \ 5 & 30 & 9 \ 6 & 62 & 10 \ \hline \end{array} \]

Analysis of Growth:

From the calculations:

  • At \( x = 3 \): \( f(3) < g(3) \)
  • At \( x = 4 \): \( f(4) > g(4) \)
  • At \( x = 5 \): \( f(5) > g(5) \)
  • At \( x = 6 \): \( f(6) > g(6) \)

Based on this, we can see that even though \( g(x) \) starts higher at \( x = 3 \), the function \( f(x) = 2^x - 2 \) grows much faster than \( g(x) = x + 4 \) as \( x \) increases.

Conclusion:

Eventually, the function \( f(x) \) will exceed \( g(x) \) and continue to grow faster as \( x \) increases. Therefore, as \( x \) gets larger, \( f(x) \) eventually exceeds \( g(x) \).